Q. The maximum volume (in $cu. m$) of the right circular cone having slant height $3\,m$ is :
Solution:
$\therefore h = 3 \cos\theta$
$ r=3 \sin\theta $
Now,
$ V = \frac{1}{3} \pi r^{2} h = \frac{\pi}{3} \left(9 \sin^{2} \theta\right).\left(3 \cos\theta\right) $
$ \therefore \frac{dV}{d\theta} = 0 \Rightarrow \sin\theta = \sqrt{\frac{2}{3}} $
Also, $ \frac{d^{2}V}{d\theta^{2}} \bigg]_{\sin\theta = \sqrt{\frac{2}{3}}} = $ negative
$\Rightarrow $ Volume is maximum,
when $ \sin\theta = \sqrt{\frac{2}{3}} $
$ \therefore V_{max } \left(\sin\theta = \sqrt{\frac{2}{3}}\right) = 2\sqrt{3}\pi$ (in cu. m)
